Empirical Spatial Ranks and Rank Scores

From Pairwise Directions to Spatial Ranks

A univariate rank tells us where an observation sits in an ordering. In several dimensions, there is no unique ordering that respects every direction. Spatial ranks instead summarize the directions from the other observations toward a point. They retain directional information while providing a bounded vector that can be transformed into a score for multivariate procedures.

The construction uses the spatial sign: for a nonzero vector, its spatial sign is its direction with length one. The spatial sign covariance and the geometry of spatial signs provide useful background, but a spatial rank averages signs over a reference distribution or sample.

The expectation defining \(R_F\) always exists because \(\|U(x-X)\|\leq 1\). The empirical definition includes the observation itself: its contribution is \(U(0)=0\), and the denominator remains \(n\). This convention makes the empirical function an average over the full empirical distribution. Some sources instead omit the self-comparison and divide by \(n-1\); formulas and bounds then differ by a factor of \(n/(n-1)\).

Direction and Magnitude

Each summand in \(R_n(x)\) points from an observed data point toward \(x\). Thus \(R_n(x)\) points in the average direction of the data relative to \(x\): a positive component indicates a net direction along that coordinate axis, while components may cancel. Its length measures the degree of directional agreement among the vectors being averaged. It is not a distance from \(x\) to the data and does not generally increase in a simple way with distance from the sample center.

At an observed point, the self-comparison is zero, so the other \(n-1\) vectors each have length one unless they also arise from coincident observations. The triangle inequality therefore gives \(\|R_{ni}\|\leq(n-1)/n\). A small rank length can arise when directions largely cancel; a larger length indicates more alignment. Neither quantity alone determines the observation's distance from the data cloud.

Basic Properties and Their Proof

The empirical ranks have useful structural properties that follow directly from their pairwise definition. In the result below, \(\mathbf{Q}\) is any orthogonal \(d\times d\) matrix, so that \(\mathbf{Q}^{\mathsf T}\mathbf{Q}=\mathbf{I}\), and \(\mathbf{a}\in\mathbb{R}^d\) is any fixed translation.

Proof. First, the spatial sign is odd: \(U(-z)=-U(z)\), including at \(z=0\). Consequently, $$ \sum_{i=1}^n R_{ni} =\frac1n\sum_{i=1}^n\sum_{j=1}^n U(\mathbf{x}_i-\mathbf{x}_j) =\frac1n\sum_{i<j}\bigl\{U(\mathbf{x}_i-\mathbf{x}_j)+U(\mathbf{x}_j-\mathbf{x}_i)\bigr\}=0. $$ For the norm bound, the \(j=i\) term is zero and every other term has norm at most one. The triangle inequality yields $$ \|R_{ni}\| \leq\frac1n\sum_{j\ne i}\|U(\mathbf{x}_i-\mathbf{x}_j)\| \leq\frac{n-1}{n}. $$ This remains true when some of those terms are zero because observations coincide.

For the transformation property, translation cancels in each difference. Orthogonality preserves lengths, so for every \(z\), \(U(\mathbf{Q}z)=\mathbf{Q}U(z)\), including \(z=0\). Thus the transformed rank at observation \(i\) is $$ \frac1n\sum_{j=1}^n U\bigl(\mathbf{Q}(\mathbf{x}_i-\mathbf{x}_j)\bigr) =\mathbf{Q}\left\{\frac1n\sum_{j=1}^n U(\mathbf{x}_i-\mathbf{x}_j)\right\} =\mathbf{Q}R_{ni}. $$ This proves all claims. \(\square\)

Connection to a Convex Distance Objective

Spatial ranks are also related to the average distance from a point to the observations. This relation explains why they can be interpreted as directional derivatives, with an important qualification when the point coincides with an observation.

Derivation. Each function \(x\mapsto\|x-\mathbf{x}_j\|\) is convex, so their average \(D_n\) is convex. If \(x\ne\mathbf{x}_j\), differentiation of the Euclidean norm gives $$ \nabla_x\|x-\mathbf{x}_j\|=\frac{x-\mathbf{x}_j}{\|x-\mathbf{x}_j\|}. $$ Summing these gradients and dividing by \(n\) gives \(\nabla D_n(x)=R_n(x)\) when no term is evaluated at zero.

At \(x=\mathbf{x}_j\), the norm term is not differentiable. To verify the subgradient claim, recall that \(g\) is a subgradient of a convex function \(D_n\) at \(x\) if \(D_n(y)\geq D_n(x)+g^{\mathsf T}(y-x)\) for every \(y\). For a nonzero \(z\), the inequality $$ \|w\|\geq\|z\|+\left(\frac{z}{\|z\|}\right)^{\mathsf T}(w-z) $$ follows from \( (z/\|z\|)^{\mathsf T}w\leq\|w\|\). For \(z=0\), the same subgradient inequality holds with subgradient zero because \(\|w\|\geq0\). Apply these inequalities with \(z=x-\mathbf{x}_j\), \(w=y-\mathbf{x}_j\), and add them over \(j\). The resulting inequality is \(D_n(y)\geq D_n(x)+R_n(x)^{\mathsf T}(y-x)\), as required.

The qualification matters: at a sample point the rank is a valid subgradient under the zero-sign convention, but it is not necessarily the gradient of \(D_n\), which may not be differentiable there.

Constructing Rank-Score Vectors

A spatial rank can be decomposed into its length and direction. A radial score function transforms the length, while the direction is retained. This gives a family of vector scores rather than a single mandatory choice.

The domain includes all empirical rank magnitudes because \(\|R_{ni}\|\leq(n-1)/n<1\). The zero convention avoids assigning a direction when the rank vector vanishes. A choice such as \(K(t)=t\) returns \(A_{ni}=R_{ni}\); a choice \(K(t)=t^2\) gives more relative weight to larger rank magnitudes. The score function is part of the procedure, not something determined by the data alone.

Why Rank Scores Matter—and a Common Pitfall

Rank-score vectors provide a way to build multivariate procedures from directions and relative rank magnitudes rather than relying directly on raw distances. Different score functions emphasize different parts of the empirical rank distribution. These vectors are ingredients used in multivariate location procedures; the choice of score function and the calibration of a particular test depend on the inferential setting.

A common mistake is to call \(\|R_{ni}\|\) the observation's distance from the sample center, or to assume it increases whenever that distance increases. The definition does not support either claim. It averages unit directions, so its magnitude records their cancellation or alignment. For example, in the one-dimensional worked example, the endpoint gaps are unequal, yet the endpoint rank magnitudes are both \(2/3\). Another mistake is to rank the magnitudes as if they were necessarily distinct ordinal ranks. Ties can occur, and the magnitudes are real-valued vector summaries rather than a guaranteed strict ordering.